Since you have that NxN is countable, it follows that (NxN)xN is countable and that Nx(NxN) is countable. Just use these facts, for any sets S and T:

If S is equinumerous with R, then SxT is equinumerous with RxT.

If S is equinumerous with R, then TxS is equinumerous with TxR.

In fact, for any natural number k, by induction, we may infer that

Nx(N...xN) (i.e. "k number of times") is equinumerous with N
and
(N...xN)xN (i.e. "k number of times") is equinmumerous with N

But isn't proving these facts the point of the exercise then? I know you know why they are correct, but does the OP?

Of course:

Spoiler:

If you have that is a set of bijections from the corresponding to the corresponding then is a bijection. Or, for just the two case that MoeBlee stated just take x_1,x_2)\mapsto (f_1(x_1),f_2(x_2))" alt="f_1\times f_2:U_1\times U_2\to V_1\times V_2x_1,x_2)\mapsto (f_1(x_1),f_2(x_2))" />

Yes, it is always a question: What theorems have been proven previous to the stated exercise?

I would think that in a set theory course (is this exercise from a set theory course?) the more basic theorem about Cartesian products would have been proven previously (and if it had not been, then it should be as soon as possible).

My point would be that there's no reason for the exercise to mention 'N' specifically when the result is general anyway, unless the intent of the exercise was just to remind the student to apply the more general result to N in particular. Moreover, as a matter of "style" I'd prefer a proof that appeals to the more general principle than to one that appeals to specific properties of N (that is, all other things being equal, better to prove something in a general way).

Anyway, the general theorem about Cartesian products is itself quite easy to prove just by putting the given bijections to their obvious use.